codeforces716c Plus and Square Root

题意：贼长，初始2，第i次操作时，可以连续加好多次i，当到达某一个完全平方数时，开更，开更后必须满足膜(i+1)==0，求每次操作需要加多少次i

题解：根据题意 a[i] + ans[i] * i = a[i+1] * a[i+1]

           ans[i] = (a[i+1] * a[i+1] - a[i]) / i

           a[i] % i == 0

           所以a[i+1] % i == 0

           可以构造a[i] = i * i * (i-1) * (i-1)

           ans[i] = (i+1)*(i+1)*i - (i-1)

int n;

int main(){
    while(scanf("%d",&n)!=EOF){
        for(int i=1;i<=n;i++){
            if(i==1)
                printf("2\n");
            else
                cout<<(LL)(i+1)*(i+1)*i-(i-1)<<endl;
        }
    }
    return 0;
}